Problem: Evaluate $\int\tan^5x\sec^4x\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac15\tan^5x+\dfrac17\tan^7x+C\\$ (Choice B) B $\dfrac16\tan^6x+\dfrac18\tan^8x+C\\$ (Choice C) C $\dfrac16\sec^6x+\dfrac18\sec^8x+C\\$ (Choice D) D $\dfrac16\sec^6x+\dfrac18\tan^8x+C\\$
Since we have an even power of $\sec x$ along with powers of $\tan x$, we keep one factor of $\sec^2x$ (to play the role of the derivative of $\tan x)$ and convert the remaining factors of $\sec^2x$ into $\tan x$ using the identity $\sec^2x=1+\tan^2x\,$. $\begin{aligned} &\phantom{=}\int\tan^5x\sec^4x\,dx \\\\ &= \int\tan^5x\sec^2x(1+\tan^2x)\,dx \\\\ &=\int\big(\tan^5x+\tan^7x)\,\sec^2x\,dx \end{aligned}$ We can now use a $u$ -substitution with $u=\tan x$ and $du = \sec^2x\,dx$. Since this is an indefinite integral, we don't have limits of integration to change, so we must convert back from $~u~~\text{to}~~\tan x~$ at the end. $\begin{aligned} &\phantom{=}\int\big(\tan^5x+\tan^7x)\,\sec^2x\,dx \\\\ &= \int \big(u^5+u^7\big)\,du \\\\ &= \dfrac{u^6}{6}+\dfrac{u^8}8+C \\\\ &=\dfrac16\tan^6x+\dfrac18\tan^8x+C \end{aligned}$